The equivalent weight of Cu 2 S in the reaction, Cu₂S + KMnO₄ → H 2 S O 4 CuSO₄ + SO₂ + MnSO₄ would be

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The equivalent weight of $Cu_{2} S$ in the reaction,

Cu₂S + KMnO₄ $\overset{H_{2} S O_{4}}{\to}$ CuSO₄ + SO₂ + MnSO₄

would be

Answer:

Solution:

The $n$ -factor of Cu₂S would be calculated as

Cu⁺¹ → Cu⁺².

For each Cu, the change in oxidation state is 1. For two Cu atoms, the change in oxidation state is 2.

S⁻² → S⁺⁴,

For one S , the change in oxidation state for one S atom is 6. The total oxidation state change is 8.

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Calculation of mole

⏱ 2mℹ️ Source: QB

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